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Homework 9
Ch22: Q 1; P 1, 3, 19; Ch24: P 57, 59
Question (Ch22):
1.
The electric field in an EM wave traveling north oscillates in an east–west plane.
Describe the direction of the magnetic field vector in this wave.
Solution
If the direction of travel for the EM wave is north and the electric field oscillates eastwest, then the magnetic field must oscillate up and down. For an EM wave, the direction
of travel, the electric field, and the magnetic field must all be perpendicular to each other.
Problems (Ch22):
*1. (II) At a given instant, a 1.8-A current flows in the wires connected to a parallelplate capacitor. What is the rate at which the electric field is changing between the
plates if the square plates are 1.60 cm on a side?
Solution
The current in the wires must also be the displacement current in the capacitor. We find the rate
at which the electric field is changing from
 E
 E 
;
I D   0 A
ID  0
;
t
 t 
2  E 
14
1.8A  8.85  1012 F m   0.0160m  
 , which gives 7.9  10 V m s.
 t 
3.
(I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at
a given point, what is the peak magnitude of the electric field?
Solution
The electric field is
E  cB   3.00  108 m s 17.5  109 T   5.25V m.
*19. (II) The magnetic field in a traveling EM wave has an rms strength of 28.5 nT. How
long does it take to deliver 235 J of energy to 1.00 cm2 of a wall that it hits
perpendicularly?
Solution
The energy per unit area per unit time is
S
cBrms 2
0
 3.00  10 m s  28.5  10 T 

 4  10 T m A 
9
8
7
2
 0.194 W m 2 .
We find the time from
 235J 
U
t

 1.21  107 s  140 days.
4
2
2
AS 1.00  10 m  0.194 W m 
Problems (Ch24):
57. (II) At what angle should the axes of two Polaroids be placed so as to reduce the
intensity of the incident unpolarized light to (a) 13 , (b) 101 ?
Solution
If the initial intensity is I 0 , through the two sheets we have
I1  12 I 0 ,
I 2  I1 cos2  ; which means
I2 1
 cos2  .
I0 2
I
(a) For 2  13 ,
I0
I
(b) For 2  101 ,
I0
1
3
1
10
 12 cos 2  gives   35.3.
 12 cos2  gives   63.4.
59. (II) Two polarizers are oriented at 38.0° to one another. Light polarized at a 19.0°
angle to each polarizer passes through both. What percent reduction in intensity
takes place?
Solution
Through the successive sheets we have
I1  I 0 cos2 1 ,
I 2  I1 cos2  2 , which gives
I 2  I 0 cos2 1 cos2  2  I 0  cos2 19.0  cos2 38.0   0.555I 0 .
Thus the reduction is 44.5%.